If \(\ts{\mathcal{T}_\alpha}\) is a family of topologies on \(X\), show that \(\bigcap \mathcal{T}_\alpha\) is a topology on \(X\). Is \(\bigcup \mathcal{T}_\alpha\) a topology on \(X\)?
Let \(\ts{\mathcal{T}_\alpha}\) be a family of topologies on \(X\). Show that there is a unique smallest topology on \(X\) containing all the collections \(\mathcal{T}_\alpha\), and a unique largest topology contained in all \(\mathcal{T}_\alpha\).
If \(X = \ts{a, b, c}\), let
\[ \mathcal{T}_1 = \ts{\varnothing, X, \ts{a}, \ts{a, b}} \quad \text{and} \quad \mathcal{T}_2 = \ts{\varnothing, X, \ts{a}, \ts{b, c}}. \]
Find the smallest topology containing \(\mathcal{T}_1\) and \(\mathcal{T}_2\), and the largest topology contained in \(\mathcal{T}_1\) and \(\mathcal{T}_2\).
Solution
1. For any family \(\{\mathcal T_\alpha\}\) of topologies on \(X\), \(\bigcap_\alpha\mathcal T_\alpha\) is a topology on \(X\).
Put \(\mathcal T=\bigcap_\alpha\mathcal T_\alpha\). Each \(\mathcal T_\alpha\) contains \(\varnothing\) and \(X\), so \(\mathcal T\) does. If \(\{U_\beta\}_{\beta\in B}\subseteq\mathcal T\), then for every \(\alpha\) each \(U_\beta\) lies in \(\mathcal T_\alpha\), so \(\bigcup_\beta U_\beta\in\mathcal T_\alpha\); hence \(\bigcup_\beta U_\beta\in\mathcal T\). If \(U_1,\ldots,U_n\in\mathcal T\), then for every \(\alpha\), \(U_1\cap\cdots\cap U_n\in\mathcal T_\alpha\); hence \(U_1\cap\cdots\cap U_n\in\mathcal T\).
2. A union of topologies on \(X\) need not be a topology on \(X\).
On \(X=\{a,b,c\}\) let \(\mathcal T_A=\{\varnothing,X,\{a\}\}\) and \(\mathcal T_B=\{\varnothing,X,\{b\}\}\). Then \(\mathcal T_A\cup\mathcal T_B=\{\varnothing,X,\{a\},\{b\}\}\) contains \(\{a\}\) and \(\{b\}\) but not \(\{a\}\cup\{b\}=\{a,b\}\).
3. \(\bigcap_\alpha\mathcal T_\alpha\) is the unique largest topology on \(X\) contained in every \(\mathcal T_\alpha\).
By step 1 it is a topology, and it is contained in every \(\mathcal T_\alpha\). If \(\mathcal T'\) is a topology with \(\mathcal T'\subseteq\mathcal T_\alpha\) for every \(\alpha\), then \(\mathcal T'\subseteq\bigcap_\alpha\mathcal T_\alpha\). A largest element of a family of sets ordered by inclusion is unique.
4. Let \(\mathfrak F\) be the set of topologies on \(X\) that contain every \(\mathcal T_\alpha\). Then \(\bigcap_{\mathcal T\in\mathfrak F}\mathcal T\) is the unique smallest topology on \(X\) containing every \(\mathcal T_\alpha\).
The discrete topology \(\mathcal P(X)\) lies in \(\mathfrak F\), so \(\mathfrak F\) is nonempty and, by step 1, \(\mathcal T_{\sup}=\bigcap_{\mathcal T\in\mathfrak F}\mathcal T\) is a topology. Each \(\mathcal T\in\mathfrak F\) contains every \(\mathcal T_\alpha\), so \(\mathcal T_{\sup}\) does, and every topology containing every \(\mathcal T_\alpha\) belongs to \(\mathfrak F\) and therefore contains \(\mathcal T_{\sup}\). A smallest element is unique.
5. In (c), the largest topology contained in \(\mathcal T_1\) and \(\mathcal T_2\) is \(\boxed{\{\varnothing,X,\{a\}\}}\).
By step 3 it is \(\mathcal T_1\cap\mathcal T_2=\{\varnothing,X,\{a\}\}\).
6. In (c), the smallest topology containing \(\mathcal T_1\) and \(\mathcal T_2\) is \(\boxed{\{\varnothing,X,\{a\},\{b\},\{a,b\},\{b,c\}\}}\).
The smallest topology containing \(\mathcal T_1\cup\mathcal T_2\) is the topology generated by the subbasis \(\{\{a\},\{a,b\},\{b,c\}\}\). Its finite intersections are \(X\), \(\{a\}\), \(\{a,b\}\), \(\{b,c\}\), \(\{a,b\}\cap\{b,c\}=\{b\}\), and \(\varnothing\), and this collection is already closed under unions.
7. Q.E.D.
Steps 1 and 2 answer (a), steps 3 and 4 prove (b), and steps 5 and 6 answer (c).